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STRUCTURE OF NEON ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the correct nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks, discovered by Gell-Mann and Zweig, in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws, which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the discovery of nuclear force and structure by applying the laws of electromagnetism.(See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Neon (Ne) possesses three stable isotopes, Ne-20, Ne-21, and Ne-22. In addition, 16 radioactive isotopes have been discovered ranging from Ne-16 to Ne-34, all short-lived. The longest-lived is Ne-24 with a half-life of 3.38 minutes. All others are under a minute, most under a second. In order to reveal the structure of neon isotopes I start with the diagram of my paper STRUCTURE OF Ne-20 AND Ne-19 . In the following diagram of the stable Ne-20 you see that the nucleons of the first horizontal plane ( HP1) have positive spins, while the nucleons of the fourth horizontal plane( HP4) have nucleons of negative spins . Particularly you see the structure of O-16 of S=0 with 16 nucleons existing from p1 to n8 (core) in which we add the n9(+1/2), the p9(-1/2) the n10(+1/2) and the p10(-1/2) giving a total S =0. Such additional nucleons make two rectangles outside the stable O-16 of S=0 and give the stable structure of Ne-20 with S =0. '' '' DIGRAM OF Ne-20 WITH S = 0 ''' '' ''p8(-1/2).n8(-1/2''' ' n7(-1/2).p7(-1/2) HP4' ' n6(+1/2).p6(+1/2).. n10(+1/2) ' ' n9(+1/2)..p5(+1/2).n5(+1/2 ) HP3 ' ' p4(-1/2).n4(-1/2).. p10(-1/2) ' ' p9(-1/2)..n3(-1/2).p3(-1/2) HP2 ' ' n2(+1/2).p2(+1/2) ' ' p1(+1/2).n1(+1/2) HP1' STRUTURE OF Ne-16 AND Ne-18 WITH S=0 ' Using the diagram of Ne-20 with S=0 we see that the absence of 4 neutrons of opposite spins (one neutron per horizontal plane) in the structure of O-16 gives the structure of Ne-16 with the same S = 0. It is similar to the structure of Ne-20 but here the small number of pn bonds of short range cannot overcome the pp repulsions of long range. For example the absent 4 neutrons of the structure of the core (O-16) as n1, n3, n6 and n8 of opposite spins reduce the np bonds and the pp repulsions lead to the decay. Similarly the absent 2 neutrons which give the N-18 like n1 and n6 of opposite spins give the same S = 0. ' ''' '''STRUCTURE OF Ne-17 WITH S =-1/2 AND Ne-19 WITH S =+1/2 Using again the diagram of Ne-20 with S =0 we see that the 3 absent neutrons of total spin S=+1/2 in the core of O-16 give the structure of Ne-17 with S=-1/2 . It is given by S = 0-(+1/2) = -1/2 . Similarly the one absent neutron of S =-1/2 in the core of O-16 gives the structure of Ne-19 with S= +1/2 . It is given by S = 0 - (-1/2)= +1/2. Note that in both cases the absence of neutrons reduces the number of pn bonds of short range which cannot overcome the pp repulsions of long range. ' ' WHY THE Ne-20 AND Ne-22 OF S=0 ARE STABLE NUCLIDES In the diagram of Ne-20 you see that the two vertical pn systems of S=0 with strong vertical bonds make two vertical rectangles of strong bonds which overcome the pp repulsions of long range. Similarly the Ne-22 of S=0 is a stable nuclide, because the protons p1 and p9 as well the protons p3 and p10 make blank positions able to receive 2n of opposite spins giving strong bonds able to overcome the pp repulsions of long range. In the diagram of Ne-22 you see that the p10(-1/2) becomes p10(+1/2) and the n10(+1/2) becomes n10(-1/2). Here you also see the 2n of opposite spins, because they exist at the first and the second horizontal plane of opposite spins. Especially we observe two bonds per neutron, because they make weak horizontal bonds and strong vertical bonds. ' DIAGRAM OF THE STRUCTURE OF Ne-22 WITH S=0' ' p8(-1/2).n8(-1/2 ' ' n7(-1/2).p7(-1/2) . .......n10(+1/2) HP4 ' ' n6(+1/2). p6(+1/2)' ' n9(+1/2)..p5(+1/2).n5(+1/2 .......p10(+1/2) HP3' ' p4(-1/2).n4(-1/2)' ' p9(-1/2)..n3(-1/2).p3(-1/2) -n HP2 ' ' n2(+1/2).p2(+1/2) ' ' +n p1(+1/2).n1(+1/2) HP1 ' STRUCTURE OF Ne-24, Ne-26, Ne-28, Ne-30, Ne-32 AND N-34 WITH S=0 ' Using the diagram of the structure of Ne-22, we see that the Ne- 24 has two more extra neutrons of opposite spins giving single bonds which lead to the beta decay. Similarly the Ne-26, Ne-28, Ne-30, Ne-32 and Ne-34 have 4, 6, 8, 10 and 12 extra neutrons of opposite spins respectively which give single bonds leading to the beta decay. ' ' ' WHY THE Ne-21 WITH S =+3/2 IS A STABLE NUCLIDE ' In the following diagram of the Ne-21 you see that it consists of the deuteron p10 n10 of S =-1 which is over the non elongated parallelepiped of S =+3 having 18 nucleons. Here the protons p10 and p7 make a blank position for receiving the extra n which makes one very strong vertical bond with p7 and a second horizontal bond with p10. That is, we have not single bonds but two bonds per neutron. Since the fourth horizontal plane has 3 nucleons of negative spin, then the total spin S=+3/2 of Ne-21 is given by S= +3 -3/2 = +3/2. Note that these two bonds per neutron of short range are able to overcome the nn repulsions of short range leading to the stability of Ne-21. ' ' ' DIAGRAM OF THE Ne-21 WITH S = +3/2 ' n10(-1/2) ' ' n p10(-1/2 -HP4' ' n7(+1/2)..........p8(+1/2)........n9(+1/2)' ' p7(+1/2) .........n8(+1/2).........p9(+1/2) +HP3' ' p2(-1/2)..........n4 (-1/2)……….p6(-1/2) ' ' n2 (-1/2)..........p4 (- 1/2)……….n6(-1/2) -HP2 ' ' n1(+1/2)..........p3(+1/2)……….n5(+1/2) ' ' p1(+1/2)..........n3(+1/2)……….p5(+1/2) +Hp1 ' STUCTURE OF Ne-25, Ne-27 AND Ne-29 WITH S = +3/2 Using the structure of Ne-21 with S =+3/2 we see that the Ne-25 of S =+3/2 has 4 extra neutrons of opposite spins which give the unstable Ne-25 because the weak single pn bonds lead to the beta decay. Similarly the Ne-27 has 6 extra neutrons of opposite spins giving single bonds which lead to the beta decay. Finally the Ne-29 with the same S =+3/2 has 8 extra neutrons of opposite spins which give 8 single bonds leading to the beta decay. ' ' ' ' STRUCTURE OF Ne-23 WITH S = +5/2 Comparing the structure of Ne-23 with that of Na-23 you can see also my diagrams of my paper STRUCTURE OF Na-23 AND Ne-23 . ' ' ' DIAGRAM OF Ne-23 WITH S = +5/2' ' p10(+1/2)n10(+1/2)' ' n9(+1/2).p9(+1/2) HP5' ' n8(-1/2).p8(-(1/2) ' ' n13(-1/2)..p7(-1/2).n7(-1/2) HP4 ' ' p6(+1/2).n6(+1/2)' ' n5(+1/2).p5 (+1/2) …..n12(+1/2) HP3' ' n4(-1/2).p4(-1/2)' ' p3(-1/2).n3(-1/2) HP2' ' p2(+1/2).n2(+1/2) ' ' n1(+1/2).p1(+1/2)…. n11(+1/2) HP1' ' ' STRUCTURE OF Ne-31 AND Ne-33 WITH S =-7/2 In the following typical diagram of the Ne-18 with S=-3 you see that it consists of a non elongated parallelepiped with 18 nucleons giving S =-3, because the third horizontal plane has negative spins (-HP3). Then adding the p10(+1/2) behind the n4(+1/2) the n10 (-1/2) behind the p8(-1/2) and the n11(-1/2) behind the p3(-1/2) we get 2 strong vertical rectangles of a total S =-1/2 .Note that these three nucleons which are not shown give a typical structure of a non elongated Ne-21 with a S=-7/2 given by S =-3 +(-1/2)= -7/2 ' '''Therefore the structure of Ne-33 with S =-7/2 has 12 extra neutrons of opposite spins which give single bonds leading to the beta decay. Similarly the Ne-33 of S =-7/2 has 14 extra neutrons of opposite spins . ' ' '''DIAGRAM OF A NON ELONGATED PARALLELEPIPED WITH 18 NUCLEONS GIVING S =-3 ' ' n7(-1/2)..........p8(-1/2)........n9(-1/2)' ' p7(-1/2) .........n8(-1/2).........p9(-1/2) -HP3' ' p2(+1/2)..........n4 (+1/2)……….p6(+1/2) ' ' n2 (+1/2)..........p4 (+1/2)……….n6(+1/2) +HP2 ' ' n1(-1/2)..........p3(-1/2)……….n5(-1/2) ' ' p1(-1/2)..........n3(-1/2)……….p5(-1/2) -Hp1 ' Category:Fundamental physics concepts